When does $(x+1)^s -x^s = x^{-s}$ have a solution? Let $0\lt s \lt 1$. I want to find the range of $s$ for which the following equation has a solution for $x\ge 0$.
$$(x+1)^s -x^s = x^{-s} $$
After looking at the graphs of the LHS and RHS and Wolfram, I believe that if $s$ is greater than some unknown value, there is always one solution. For example, WolframAlpha gives no solutions for $s=0.5$ but one for $s=0.6$.
How can I obtain this transition point? Does its closed-form even exist?
I might add an idea. At that special value of $s$, the two curves must touch each other, giving the system $$(x+1)^s -x^s = x^{-s} \\ (x+1)^{s-1} -x^{s-1} = -x^{-s-1} $$
Maybe there’s a way to solve for $s$.
 A: I believe it fails at $s=\frac 12$ but will succeed at any $s$ above that.  We have
$\sqrt{x+1}-\sqrt x \approx \frac 12 x^{-1/2}-\frac 18x^{-3/2}$ as $x \to \infty$  The leading factor of $\frac 12$ says there will be no solution because the left side is always larger. For $x=0.52$ I found a solution around $x=1.6\cdot 10^7$ but closer than that I am running into precision errors in my spreadsheet.  I tried to get Alpha to expand the $s=0.51$ case but it wouldn't.
A: I prefer to add another answer just dedicated to numerical aspects.
For a given value of $s$, instead of solving for $x$
$$f(x)=(x+1)^s -x^s - x^{-s}=0$$ it looks much better to solve instead
$$\color{red}{h(x)=\log \left( (x(x+1))^s-x^{2 s}\right)=0}$$
Using what was proposed in the previous answer
$$x_0=\exp\left(\frac{\log (10)}{\sqrt{3}} \,\frac{s-1}{1-2s} \right)$$ is a good starting point for Newton method (it could probably be significantly improved).
For $s=0.52$ used by @Ross Millikan, the iterates are
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 8.47624\times 10^6 \\
 1 & 1.18276\times 10^7 \\
 2 & 1.25635\times 10^7 \\
 3 & 1.25868\times 10^7 
\end{array}
\right)$$
Still with the idea of finding a better conditionning, it is definitrly better to let $\color{red}{x=e^t}$; (his makes $f(x)$ much more linear. For the same problem as above with the same initial conditions, Newton iterates are
$$\left(
\begin{array}{cc}
n & t_n \\
 0 & 15.952777 \\
 1 & 16.348162 
\end{array}
\right)$$
For $s=0.51$ which was making problems
$$\left(
\begin{array}{cc}
 n & t_n \\
 0 & 32.570254 \\
 1 & 33.667228 
\end{array}
\right)$$
Now, a much better estimate
$$t_0=\frac{764}{583} \theta+\frac{6}{1015}\theta^2-\frac{1}{7097}\theta^3 \qquad \text{with} \qquad \theta=\frac{s-1}{1-2 s}
$$
For the two above examples, the estimates are $16.3333 $ and $33.5826$.
